how about let u = 2x-3
then 2x = u + 3
x = (u+3)/2
then y = 2(u+3)^2 /4 - 2
= (1/2)(u^2 + 6y + 9) - 2
dy/du = (1/2)(2u + 6) = u + 3
= (2x-3) + 3 = 2x
dy/d(2x-3) = 2x
If y=3x^2-2, then dy/d(2x-3)= ?
I've never taken derivatives with respect to anything other than a single variable, such as x, y or t, so i'm not sure how to start this problem! Thanks
3 answers
or, you can let
u = 2x-3
dx/du = 1/2
dy/du = dy/dx * dx/du
= (6x)(1/2) = 3x
Reiny: there's a typo in your 2nd paragraph:
y = 3(u+3)^2-2
u = 2x-3
dx/du = 1/2
dy/du = dy/dx * dx/du
= (6x)(1/2) = 3x
Reiny: there's a typo in your 2nd paragraph:
y = 3(u+3)^2-2
Thanks Steve, I see it now, used 2x^2 instead of 3x^2
1 answer